Error Invalid Type Argument Of Unary Have Ong Int

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Error Invalid Type Argument Of Unary ‘*’ (have ‘double’)

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Error Invalid Type Argument Of Unary '*' (have 'char')

them; it only takes a minute: Sign up error: invalid type argument of ‘unary *’ (have ‘int’) up vote 20 down vote favorite 7 I have a C Program: #include int main(){ int b = 10; //assign the invalid type argument of unary ‘*’ (have ‘int’) c integer 10 to variable 'b' int *a; //declare a pointer to an integer 'a' a=(int *)&b; //Get the memory location of variable 'b' cast it //to an int pointer and assign it to pointer 'a' int *c; //declare a pointer to an integer 'c' c=(int *)&a; //Get the memory location of variable 'a' which is //a pointer to 'b'. Cast that to an int pointer //and assign it to pointer 'c'. printf("%d",(**c)); //ERROR HAPPENS HERE. return 0; } Compiler invalid type argument of unary ‘*’ (have ‘float’) produces an error: error: invalid type argument of ‘unary *’ (have ‘int’) Can someone explain what this error means? c pointers share|improve this question edited Sep 30 '13 at 19:23 Eric Leschinski 46.4k23221190 asked Mar 28 '11 at 7:30 picstand 112127 add a comment| 4 Answers 4 active oldest votes up vote 14 down vote accepted Since c is holding the address of an integer pointer, its type should be int**: int **c; c = &a; The entire program becomes: #include int main(){ int b=10; int *a; a=&b; int **c; c=&a; printf("%d",(**c)); //successfully prints 10 return 0; } share|improve this answer edited Sep 30 '13 at 19:24 Eric Leschinski 46.4k23221190 answered Mar 28 '11 at 7:41 codaddict 251k50362442 3 Also note the lack of casts in the answer. The casts in the question hide the problems on the line that is assigning an int ** to a int *. (c=(int *)&a;) –Thanatos Sep 30 '13 at 22:54 add a comment| up vote 8 down vote Barebones C program to produce the above error: #include using namespace std; int main(){ char *p; *p = 'c'; cout << *p[0]; //error: invalid type argument of `unary *' //peeking too deeply into p, that's a paddlin. cout << **p; //error: invalid type argument of `unary *' //peeking too deeply into p, you better believe that's a paddlin. } ELI5: You have a big plasma T

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Invalid Type Argument Of Unary '*' (have 'float')

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Invalid Type Argument Of Unary ‘*’ (have ‘char’)

Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; http://stackoverflow.com/questions/5455866/error-invalid-type-argument-of-unary-have-int it only takes a minute: Sign up error: invalid type argument of unary '*' up vote 2 down vote favorite I don't understand these errors can someone explain? error: invalid type argument of unary '' (have 'double') error: invalid type argument of unary '' (have 'double') error: invalid type argument of unary '*' (have 'double') double getMedian(double *array, int http://stackoverflow.com/questions/22902985/error-invalid-type-argument-of-unary *hours){ if (*hours <= 0) return 0; if (*hours % 2) return (float)*array[(*hours + 1) / 2]; else{int pos = *hours / 2; return (float)(*array[pos] + *array[pos + 1]) / 2;}} c++ arrays pointers share|improve this question edited Apr 14 at 11:19 Slothworks 468514 asked Apr 7 '14 at 3:10 user3502479 2525 add a comment| 3 Answers 3 active oldest votes up vote 4 down vote accepted You are already dereferencing array with the [] operator. What you want is: double getMedian(double *array, int *hours){ if (*hours <= 0) return 0; if (*hours % 2) return (float)array[(*hours + 1) / 2]; else{int pos = *hours / 2; return (float)(array[pos] + array[pos + 1]) / 2;}} Note that writing x[y] is shorthand for *(x + (y)). In your code, you have essentially have the equivalent of **array. share|improve this answer answered Apr 7 '14 at 3:19 ppl 1,004516 add a comment| up vote 1 down vote When you use the [] operator on the arrays or pointers, you don't have to dereference them again to get the valu

question and get tips & solutions from a community of 418,513 IT Pros & Developers. It's quick https://bytes.com/topic/c/answers/902232-invalid-type-argument-unary & easy. invalid type argument of 'unary*' P: 1 Stephanie Smith I https://bytes.com/topic/c/answers/950223-error-invalid-type-argument-unary-have-int always get this error message! Here's the code: Expand|Select|Wrap|Line Numbers #include #include intmain(void) { char*string; charlength_string=0; charbuffer_for_string=0; char*temp_one=malloc(sizeof(char)); if(temp_one==NULL) { free(string); return-2; } else { string=temp_one; temp_one=NULL; } while((buffer_for_string=getchar())!=EOF) { **string=buffer_for_string;//errorappearshere char*temp_two=realloc(*string,sizeof(char)*(length_string+1)); if(temp_two==NULL) { free(string); return-2; } else { string=temp_two; temp_two=NULL; } length_string++; } getchar(); return0; invalid type } What did I do wrong? Thanks! Nov 17 '10 #1 Post Reply ✓ answered by donbock A variable is referenced directly via its name. A variable is referenced indirectly via some other variable that contains a pointer to the first variable. Indirect references are accomplished through the dereference operator (asterisk, *). For example: Expand|Select|Wrap|Line Numbers inta;//Declareint. int*p;//Declarepointer-to-int. a=10;//Directreferenceto'a'. invalid type argument p=&a;//Set'pa'topointat'a'. *p=20;//Indirectreferenceto'a'. Line 5 is said to dereference pointer 'p'. The compiler supports more levels of indirection than most people should use. It is uncommon (but not unusual) for me to use two levels of indirection; but I rarely use three levels of indirection and I don't remember ever using four. For example: Expand|Select|Wrap|Line Numbers inta;//Declareint. int*p;//Declarepointer-to-int. int**pp;//Declarepointer-to-pointer-to-int. a=10;//Directreferenceto'a'. pp=&p;//Set'pp'topointat'p'. *pp=&a;//Indirectlyset'p'topointat'a'. **pp=20;//[double]Indirectreferenceto'a'. *p=30;//Indirectreferenceto'a'. Lines 6 and 7 are said to dereference pointer 'pp'. Line 8 is said to dereference pointer 'p'. Lines 23 and 24 of your original program do not use the dereference operator properly. Note: this post does not show all of the ways to declare and dereference pointers. Share this Question 2 Replies Expert Mod 100+ P: 931 Meetee Why have you taken two pointers there? Please check this for reference. Nov 17 '10 #2 reply Expert 100+ P: 2,295 donbock A variable is referenced directly via its name. A variable is referenced indirectly via some other variable that contains a pointer to the first variable. Indirect references are accomplished through the dereference oper

help? Post your question and get tips & solutions from a community of 418,513 IT Pros & Developers. It's quick & easy. error: invalid type argument of unary ‘*’ (have ‘int’) P: 1 adimustdie Expand|Select|Wrap|Line Numbers char*chktype(char*Buffer,intSize)//checkstheContent-Typeandgivesextension { unsignedshortiphdrlen; structiphdr*iph=(structiphdr*)(Buffer+sizeof(structethhdr)); iphdrlen=iph->ihl*4; structtcphdr*tcph=(structtcphdr*)(Buffer+iphdrlen+sizeof(structethhdr)); intheader_size=sizeof(structethhdr)+iphdrlen+tcph->doff*4; char*strng="Content-Type:"; size_tsz; char*found=strstr(Buffer+header_size,strng); char*found1=strstr(found,"\r\n"); sz=strlen(found)-strlen(found1); char*type=malloc(sz+1); strncpy(type,found,sz); if(strncmp("Content-Type:text/html",type,23)==0) { return(".html"); } else { return(""); } } i am calling this function in the main as Expand|Select|Wrap|Line Numbers char*extension=chktype(buffer,size); printf("%s\r\n",extension); but it gives me error Expand|Select|Wrap|Line Numbers Dummy.c:74:9:error:invalidtypeargumentofunary‘*’(have‘int’) Dummy.c:76:5:warning:passingargument1of‘chktype’discards‘const’qualifierfrompointertargettype[enabledbydefault] Dummy.c:22:7:note:expected‘char*’butargumentisoftype‘constu_char*’ Dummy.c:78:5:warning:passingargument1of‘chktype’discards‘const’qualifierfrompointertargettype[enabledbydefault] Dummy.c:22:7:note:expected‘char*’butargumentisoftype‘constu_char*’ sorry for bad english, i am new to C, can anyone help. `buffer` is probly `const u_char *` type is this related in any way?? what is my mistake?? how can i get rid of error.? Jun 27 '13 #1 Post Reply Share this Question 1 Reply Expert Mod 5K+ P: 8,905 weaknessforcats The type of your calling value has to match the type the function is expecting. The function expects a char* but you say you are probably using a u_char*. That, in itself, won't work. Jun 28 '13 #2 reply Message Cancel Changes Post your reply Join Now >> Sign in to post your reply or Sign up for a free account. Similar topics invalid type argument of 'unary*' invalid type argument of ‘unary *’ (have ‘int’) SFINAE -- basis: array of void is invalid type invalid operands to binary == or wrong type argument to unary exclamation mark invalid lvalue in unary `&' invalid lvalue in unary '&' .... why LKM invalid type argument of '->' pointer-to-pointer (invalid lvalue in unary `&) Compile error: invalid type modifier within pointer declarator Questions about defaut template type argument Browse more C / C++ Questions on Bytes Question stats viewed: 5700 repli